Question

Prove that SinAsin (60-A)sin(60+A)=1/4sin3A

Answer 1

Solution:-
L.H.S. = sin A sin (60° - A) sin (60° + A)
= sin A [ sin² 60° - sin² A]      {sin (A + B) sin (A - B) = sin² A - sin² B }
= sin A [3/4 - sin² A]
= (3sin A)/4 - sin³ A = 1/4 [3sin A - 4sin³ A]
= 1/4sin³ A = R.H.S.
Hence proved.

Answer 2

Step-by-step explanation:

sinA.sin(60+A).sin(60-A)

=sinA(sin²60-sin²A)

=sinA(3/4-sin²A)

=3sinA/4-sin³A

=1/4(3sinA-4sin³A)

=1/4sin3A